1. Neutralization
Chapter 21

2. Neutralization Reactions
A reaction between a strong acid and a strong base in an aqueous solution that produces a salt and water- a salt is an ionic compound
These are double replacement reactions
Used to produce pure salts
HCl KOH KCl + H2O
The water is evaporated leaving behind the salt KCl

3. Titrations-
used to calculate the unknown concentration of an acid or base when you know the concentration of the other.
Standard solution-
solution of known concentration
End Point-
the point in the titration where the indicator changes color

4. From a balanced equation, you can determine how many moles of a substance is needed to react with another substance to be neutralized:
EXAMPLE: How many moles of sulfuric acid are needed to neutralize 0.5 mol of sodium hydroxide?
H2SO NaOH Na2SO H2O
Using stoichiometry:
0.5 mol NaOH x 1 mol H2SO4
2 mol NaOH
= 0.25 mol H2SO4
Do 1 & 2 page 616

5. Stoichiometry can be used to determine the strength of an unknown acid or base:
Titrations: Steps:
A measured volume of unknown concentration acid or base is added to a flask
Several drops of the indicator is added to the solution
Measured volume of a base or acid of known concentration are mixed into the flask until the indicator just barely changes color

6. EXAMPLE: A 25 mL solution of H2SO4 is completely neutralized by 18 mL of 1.0 M NaOH using phenolphthalein as an indicator. What is the concentration of the H2SO4solution? H2SO4 + 2NaOH Na2SO4 + 2H2O L NaOH x 1.0 mol NaOH x 1 mol H2SO4 1 L NaOH 2 mol NaOH = mol H2SO L = 0.36 M

7. Equivalents-
the amount of acid or base that will give 1 mole of hydrogen or hydroxide
Examples:
HNO3 H NO3-
1 mole of HNO3 gives 1 mole H+
1 mole HNO3 is 1 equivalent of HNO3
H2SO4 2H SO42-
1 mole of H2SO4 is 2 equivalents of H2SO4

8. Gram Equivalent Mass-
the mass of one equivalent of a substance
Molar Mass = gram equiv. mass
# equiv/mol
Example:
HCl = 36.5/1 equiv/mol = 36.5 g/equiv
H2SO4 = 98 g/mol/2 equiv/mol = 49 g/equiv

9. Normality-
concentration of a solution expressed as the number of equivalents of solute in 1 liter of solution
N = equiv / L
Equivalence point-
the point of neutralization in a titration
The number of equivalents of acid and base are equal at this point so:
NA x VA = NB x VB